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I'm currently studying complex analysis, and I've found two differnent definitions of a smooth path. Here's one defined as in Beck's book:

A path $\gamma$ is smooth if it's continuously differentiable and has nonzero derivatives.

However, some authors define a smooth path without requiring $\gamma$ to have nonzero derivatives. I'm just wondering why would I need the condition of having nonzero derivatives, since being continuously differentiable already ensures the existence of the contour integral. Could anyone give some examples to illustrate the point? Thanks.

Nazono Sumiko
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    A derivative being zero allows for sharp corners, which would not be smooth, but is still a fine contour. – Ninad Munshi Jul 11 '21 at 08:39
  • A simple example of a sharp corner is $\gamma(t)=t^2+it^3$ for $-1\leq t\leq1$. – Andreas Blass Jul 11 '21 at 12:02
  • @Andreas Blass yeah,but I don't see any necessity to add this requirement, since in many applications we allowed the contour to have corners. – Nazono Sumiko Jul 12 '21 at 11:10
  • Piecewise smooth curves work fine as contours. Integration theory doesn't get into trouble until curves have cusps (e.g. like the one in my example) or worse all over the place. My first comment was only about the actual question --- illustrating the difference between "continuously differentiable" (but possibly zero derivatives) and the intuitive idea of "smooth curve". – Andreas Blass Jul 12 '21 at 13:31

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